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题解

博弈论的前提是双方都是绝顶聪明的人

所以这种题目显然不是我应该做的

题解

//minamoto

#include<bits/stdc++.h>

#define R register

#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)

#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)

#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)

using namespace std;

char buf[1<<21],*p1=buf,*p2=buf;

inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}

int read(){

    R int res,f=1;R char ch;

    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);

    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');

    return res*f;

}

int read(char *s){

	R int len=0;R char ch;while(((ch=getc())>'Z'||ch<'A'));

	for(s[++len]=ch;(ch=getc())>='A'&&ch<='Z';s[++len]=ch);

	return s[len+1]='\0',len;

}

const int N=1e6+5;

struct eg{int v,nx;}e[N<<1];int head[N],deg[N],tot;

inline void add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot,++deg[v];}

char s[N];int n;

void solve(){

	n=read(),tot=0;

	memset(head,0,4*(n+1)),memset(deg,0,4*(n+1));

	for(R int i=1,u,v;i<n;++i)u=read(),v=read(),add(u,v),add(v,u);

	read(s);

	if(n<3)return puts("Draw"),void();

	if(n==3){

		int cnt=(s[1]=='W')+(s[2]=='W')+(s[3]=='W');

		puts(cnt>1?"White":"Draw");

		return;

	}

	fp(i,1,n)if(s[i]=='W')head[++n]=0,add(i,n),add(n,i),deg[n]=3;

	int cc=0;

	fp(u,1,n)if(deg[u]>3)return puts("White"),void();

	else if(deg[u]==3){

		int cnt=0;

		go(u)cnt+=deg[v]>=2;

		if(cnt>=2)return puts("White"),void();

		++cc;

	}

	puts((cc==2&&(n&1))?"White":"Draw");

}

int main(){

//	freopen("testdata.in","r",stdin);

	int T=read();

	while(T--)solve();

	return 0;

}